luoguP3224 [HNOI2012]永无乡

https://www.luogu.org/problemnew/show/P3224

考虑对每个岛维护一颗平衡树，用并查集维护连通性，启发式合并即可

这东西其实是一个大暴力，每次把节点少的平衡树合并到节点多的平衡树里

这样可以保证每个点合并一次树的大小*2，每个点最多被插入 log(n) 次，复杂度正确

我使用了简单好写的 leafy tree 作为平衡树，不喜勿喷

#include <bits/stdc++.h>
#define update(u) if(u -> left -> size) u -> size = u -> left -> size + u -> right -> size, u -> value = u -> right -> value
#define new_Node(a, b, c, d) (&(*st[cnt++] = Node(a, b, c, d)))
#define merge(a, b) new_Node(a -> size + b -> size, b -> value, a, b)
#define ratio 4
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
template <typename _T>
inline void read(_T &f) {
    f = 0; _T fu = 1; char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') fu = -1; c = getchar();}
    while(c >= '0' && c <= '9') {f = (f << 3) + (f << 1) + (c & 15); c = getchar();}
    f *= fu;
}
const int N = 300000 + 10;
struct Node {
	int size, value;
	Node *left, *right;
	Node (int a, int b, Node *c, Node *d) : size(a), value(b), left(c), right(d) {}
	Node () {}
}*root[N], *st[N], t[N], *null;
int w[N], v[N], f[N], pre[N], len;
int n, m, q, cnt = 0;
void maintain(Node *u) {
	if(u -> left -> size > u -> right -> size * ratio) u -> right = merge(u -> left -> right, u -> right), st[--cnt] = u -> left, u -> left = u -> left -> left;
	if(u -> right -> size > u -> left -> size * ratio) u -> left = merge(u -> left, u -> right -> left), st[--cnt] = u -> right, u -> right = u -> right -> right;
}
void ins(Node *u, int x) {
	if(u -> size == 1) u -> left = new_Node(1, min(u -> value, x), null, null), u -> right = new_Node(1, max(u -> value, x), null, null);
	else ins(x > u -> left -> value ? u -> right : u -> left, x);
	update(u); maintain(u);
}
int find(Node *u, int x) {
	if(u -> size == 1) return u -> value;
	return x > u -> left -> size ? find(u -> right, x - u -> left -> size) : find(u -> left, x);
}
void dfs(Node *u) {
	if(u == null) return;
	st[--cnt] = u;
	dfs(u -> left);
	if(u -> size == 1) w[++len] = u -> value;
	dfs(u -> right);
}
int Merge(int a, int b) {
	if(root[a] -> size < root[b] -> size) swap(a, b);
	len = 0; dfs(root[b]);
	for(register int i = 1; i < len; i++) {
		ins(root[a], w[i]);
	}
	return a;
}
int find(int x) {
	return f[x] == x ? x : f[x] = find(f[x]);
}
int main() {
	read(n); read(m);
	null = new Node(0, 0, 0, 0);
	for(register int i = 0; i <= 300000; i++) st[i] = &t[i];
	for(register int i = 1; i <= n; i++) root[i] = new Node(1, INT_MAX, null, null), read(v[i]), ins(root[i], v[i]), f[i] = i, pre[v[i]] = i;
	for(register int i = 1; i <= m; i++) {
		int a, b;
		read(a); read(b);
		int x = find(a), y = find(b);
		if(x == y) continue;
		int fa = Merge(x, y);
		f[x] = f[y] = fa;
	}
	read(q);
	for(register int i = 1; i <= q; i++) {
		char c = getchar();
		while(c != 'Q' && c != 'B') c = getchar();
		if(c == 'Q') {
			int a, b;
			read(a); read(b);
			int x = find(a);
			if(root[x] -> size - 1 < b) {
				printf("-1\n");
				continue;
			}
			printf("%d\n", pre[find(root[x], b)]);
		} else {
			int a, b;
			read(a); read(b);
			int x = find(a), y = find(b);
			if(x == y) continue;
			int fa = Merge(x, y);
			f[x] = f[y] = fa;
		}
	}
	return 0;
}